Proving triangle inequality involving measures and symmetric differences

Question

I encountered the following exercise in a measure theory text:

"Prove the triangle inequality $\mu(A \Delta C)\leq \mu(A \Delta B) + \mu(B \Delta C).$ Hint: Note that $1_{A\Delta B}=|1_{A}-1_{B}|.$"

Now I was able to show this using a Venn diagram and breaking up and recombining some subsets, but this method did not involve any indicator functions as the hint suggested.

Could anyone suggest any idea of what was the approach that the hint was referring to?

Thank you.

Answer 1

The identity itself $1_{A \triangle B} = |1_A - 1_B|$ can be proved with some casework similar to a Venn diagram.

To utilize the hint,

• note that $$1_{A \triangle C} = |1_A - 1_C| = |1_A - 1_B + 1_B - 1_C| \le |1_A - 1_B | + |1_B - 1_C| = 1_{A \triangle B} + 1_{B \triangle C}$$
• note that for a measurable set the quantitty $\mu(E) $ can be written as $\mu(E)=\int 1_E \, d\mu$.

Answer 2

Recall that the measure of a measurable set is the $\mu$-integral of the set indicator. By following the hint we get $$\mu(A\Delta C)=\int\mathbb{I}_{A \Delta C}(x)\mu(dx)=\int|\mathbb{I}_{A}(x)-\mathbb{I}_C(x)|\mu(dx)$$ By triangle inequality, pointwise, $$|\mathbb{I}_{A}(x)-\mathbb{I}_C(x)|\leq|\mathbb{I}_{A}(x)-\mathbb{I}_B(x)|+|\mathbb{I}_{B}(x)-\mathbb{I}_C(x)|$$ And finally the properties of the $\mu$-integral of positive measurable functions $$\mu(A \Delta C) \leq \mu(A \Delta B) + \mu(B\Delta C)$$